Help finding a summation using CAS

Question

While approximating an integral by midpoint rule, I ended up with $$\iint_R f(x, y)\hspace{1mm}dA\approx \dfrac{1}{n^2}\sum_{i=0}^{n-1}\sum_{j=0}^{n-1} f\left[\dfrac{1}{2n}+\dfrac{i}{n},\hspace{3mm} \dfrac{1}{2n}+\dfrac{j}{n}\right]\hspace{1mm} $$

Where $f(x, y) = \sin(x+\sqrt{y})$

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Can any one please find this using some type of CAS, I don't know how to use the advanced calculating tools.

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I will be thankful if you could find the sum for $n=1, 2, 4, 8, 16, 32$

Answer 1

In Mathematica and/or WolframAlpha:

Input: "(1/64^2)*Sum[Sum[Sin[(i+0.5)/64+Sqrt[(j+0.5)/64]],{j,0,63}],{i,0,63}]"

Output: "0.857956"

You can replace the 64 and 63 with whatever value of n and n-1 you want.

Also, using the substitution $y = u^2$, we can compute the exact value of the integral:

$\displaystyle\int_0^1\int_0^1\sin(x+\sqrt{y})\,dx\,dy = 4\cos 1 + 2\sin 1 - 2\cos 2 - 2\sin 2 - 2 \approx 0.857850$.

Answer 2

In Maple:

[> f := (x,y)-> sin(x+sqrt(y)):
   for k from 0 to 5 do evalf(1/2^(2*k)*Sum(Sum(f(1/(2*2^k)+i/2^k, 1/(2*2^k)+j/2^k), i = 1 .. 2^k-1), j = 1 .. 2^k-1), 5) end do;

                        0.84016
                        0.84016
                        0.84016
                        0.84016
                        0.84016
                        0.84016